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%%文档的题目、作者与日期
\author{王立庆（2020级数学与应用数学1班） }
\title{随机分析入门习题解答 -- 布朗运动}
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%\date{2021 年 9 月 14 日}
%\date{March 9, 2021}

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\maketitle

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\begin{enumerate}\itemsep1em

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\item  %第1题：多项选择
A Stochastic process $B=\{B_t,\, t\in [0,\infty)\}$ is called standard Brownian motion or a Wiener process if the following conditions are satisfied.

\begin{enumerate}
\item[a.]  It starts at zero and it has continuous sample paths. 
\item[b.]  It has stationary increments, i.e., for all $t>s\ge 0$, and $h\ge 0$, the increment $X_t - X_s$ and $X_{t+h} - X_{s+h}$ have the same distribution. 
\item[c.]  It has independent increments, i.e., for all $n\ge 1$ and $t_n > \cdots > t_2 > t_1 \ge 0$, 
the increments $X_{t_2} - X_{t_1},  \cdots, X_{t_n} - X_{t_{n-1}}$ are independent random variables. 
\item[d.]  For every $t>0$, $B_t$ has a normal distribution $N(0,t)$. 
\end{enumerate} 

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{\color{red}解答：abcd. 一个连续时间随机过程如果符合这些条件，那么称为标准布朗运动。

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\item  %第2题 
$B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Find the distribution of the random variable $B_5-B_3$. 
\begin{enumerate}
\item[a.]  $N(0,2)$. 
\item[b.]  $N(0,3)$. 
\item[c.]  $N(0,5)$. 
\item[d.]  $N(0,8)$. 
\end{enumerate} 

\vspace{0.2cm}

{\color{red}解答：a. 增量 $B_5-B_3$ 与随机变量 $B_2$ 有同样的分布，即均值为零，方差为 2 的正态分布。

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\item  %第3题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Find the distribution of the random variable $B_5+B_3$. 
\begin{enumerate}
\item[a.]  $N(0,3)$. 
\item[b.]  $N(0,5)$. 
\item[c.]  $N(0,8)$. 
\item[d.]  $N(0,14)$. 
\end{enumerate} 

\vspace{0.2cm}

{\color{red}解答：d. 增量 $B_5$ 与 $B_3$ 不是独立的。因此首先将所求随机变量写成 $$B_5+B_3 = (B_5-B_3) + 2B_3,$$
其中根据布朗运动的独立增量性质，随机变量 $B_5-B_3$ 与 $B_3$ 是独立的。因为 $B_5-B_3\sim N(0,2)$, 而 $B_3\sim N(0,3)$ 故 $2B_3\sim N(0,12)$, 因此  $(B_5-B_3) + 2B_3$ 服从正态分布 $N(0,14)$. 

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\item  %第4题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Find the expectation function $\mu(t)=\mathbb{E}(B_t)$ and covariance function $c(t,s)=\mathbb{E}(B_tB_s)$. Here $t<s$. 
\begin{enumerate}
\item[a.]  $\mu(t)=0,\,\, c(t,s)=t$. 
\item[b.]  $\mu(t)=t,\,\, c(t,s)=t$. 
\item[c.]  $\mu(t)=0,\,\, c(t,s)=s$. 
\item[d.]  $\mu(t)=t,\,\, c(t,s)=s$. 
\end{enumerate} 

\vspace{0.2cm}

{\color{red}解答：b. 从 $B_t\sim N(0,t)$ 我们立刻知道均值函数 $\mu(t)=0$. 为计算协方差函数 $\mathbb{E}(B_tB_s)$, 因为 $t<s$, 所以我们首先将 $B_tB_s$ 写成 $B_t[B_t+(B_s-B_t)]$. 因为 $B_t$ 与 $B_s-B_t$ 相互独立，所以 $\mathbb{E}[B_t(B_s-B_t)] = 0$. 所以协方差函数为 $\mathbb{E}(B_t^2)=t.$

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\item  %第5题 
Brownian motion is $0.5$-self-similar, i.e., for all $n\ge 1$ and $t_n > \cdots > t_2 > t_1 \ge 0$, for all $T>0$, 
the vector of random variables $\left( \sqrt{T}B_{t_1}, \sqrt{T}B_{t_2}, \cdots, \sqrt{T}B_{t_n} \right)$ has the same distribution as the vector of random variables $\left( B_{Tt_1}, B_{Tt_2}, \cdots, B_{Tt_n} \right)$. Find the correct statements. 
\begin{enumerate}
\item[a.]  The random variables $2B_2$ and $B_8$ have the same distribution. 
\item[b.]  The random vectors $(2B_1, 2B_2, 2B_3)$ and $(B_4, B_8, B_{12})$ have the same distribution. 
\item[c.]  The random variables $3B_1$ and $B_9$ are equal.  
\item[d.]  The sample path of a Brownian motion is nowhere differentiable. 
\end{enumerate} 

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{\color{red}解答：abd. 随机变量 $3B_1$ 与 $B_9$ 有相同的分布，即 $N(0,9)$, 但是它们不相等。


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\item  %第6题 
The variation of a function $f(t),0\le t\le T$ is defined as the supreme of the variations $$\sup_\tau \sum\limits_{k=1}^{n} \Big{|} f(t_k) - f(t_{k-1}) \Big{|}, $$
where the supremum is taken over all possible partitions $\tau: \,\, 0=t_0<t_1<\cdots<t_n=T$. 
Find the correct statements about variations of functions. 
\begin{enumerate}
\item[a.]  The variation of the function $f(t)=t^2$ over $[0,2]$ is $4$.
\item[b.]  The variation of the function $f(t)=t^2$ over $[-2,2]$ is $4$.
\item[c.]  The variation of the function $f(t)=\sin(t)$ over $[0,2\pi]$ is $4$.
\item[d.]  The variation of the Brownian motion $B_t$ over $[0,T]$ is infinity for any $T>0$. 
\end{enumerate} 

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{\color{red}解答：acd. 对于单调递增或单调递减的函数 $f(t),t\in [a,b]$, 其变差为 $|f(b)-f(a)|$. 因此 $f(t)=t^2$ 在区间 $[0,2]$ 上的变差是 $4$, 在区间 $[-2,2]$ 上的变差是 $8$. 正弦函数 $\sin(x)$ 在区间 $[0,\frac{\pi}{2}]$ 上的变差是 $1$, 在区间 $[\frac{\pi}{2}, \pi]$ 上的变差也是 $1$, 如此可以计算在区间 $[0,2\pi]$ 上的变差为 $4$. 另一方面，布朗运动的路径在任何（长度大于零的）区间上的变差是无穷大。

The unbounded variation and non-differentiability of Brownian sample paths are major reasons for the failure of classical integration methods, when applied to these paths, and for the introduction of stochastic calculus. 


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\item  %第7题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Consider the process $$X_t = B_t -tB_1, \,\, 0\le t\le 1. $$
Then $X_0=0, X_1=0$. For this reason this process has the name Brownian bridge. Find the correct statements. 
\begin{enumerate}
\item[a.]  The finite dimensional distributions of $X$ are multivariate Gaussian. 
\item[b.]  The expectation function of $X$ is $\mu(t)=0$. 
\item[c.]  The covariance function of $X$ is $c(t,s)=\min(t,s)-ts$, $s,t\in [0,1]$. 
\item[d.]  The Brownian bridge appears as the limit process of the normalized empirical distribution function of a sample of uniform random variables. 
\end{enumerate} 

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{\color{red}解答：abcd. 协方差函数需要计算。设 $t<s$, 于是 
\begin{eqnarray*}
c(t,s) &=& \mathbb{E}(X_tX_s) \\
&=& \mathbb{E}[(B_t -tB_1)(B_s -sB_1)] \\ 
&=& \mathbb{E}(B_tB_s -tB_1B_s - sB_1B_t + tsB_1^2) \\
&=& t - ts - st +ts \\
&=& t - ts. 
\end{eqnarray*}

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\item  %第8题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. 
Consider the following two processes $Y_t=\mu t + \sigma B_t$ and $X_t = \exp(\mu t + \sigma B_t)$. 
Find the correct statements.
\begin{enumerate}
\item[a.]  The process $Y$ has the name Brownian motion with drift. 
\item[b.]  This process $X$ has the name Geometric Brownian motion. 
\item[c.]  Compute the expectation function of the process $X$, we see that 
$\mu(t) = \exp\left( \mu t+\sigma^2 t \right).$
\item[d.]  Compute the expectation function of the process $X$, we see that 
$\mu(t) = \exp\left( \mu t+\frac{1}{2}\sigma^2 t \right).$
\end{enumerate} 

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{\color{red}解答：abd. 对于标准正态分布 $Z$, 我们可以计算随机变量 $\exp(\lambda Z)$ 的数学期望
\begin{eqnarray*}
\mathbb{E}\left[ \exp(\lambda Z) \right] = \exp\left( \frac{1}{2}\lambda^2\right). 
\end{eqnarray*}

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\end{enumerate}

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